if A S, then X \ A S, and if (A n ) n is a sequence of sets in S, then n A n S,


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1 Lecture 5: Borel Sets Topologically, the Borel sets i a topological space are the σalgebra geerated by the ope sets. Oe ca build up the Borel sets from the ope sets by iteratig the operatios of complemetatio ad taig coutable uios. This geerates sets that are more ad more complicated, which is refelcted i the Borel hierarchy. The complexity is reflected o the logical side by the umber of quatifier chages eeded to defie the set. There is a close coectio betwee the arithmetical ad the Borel hierarchy. Defiitio 5.1: Let X be a set. A σalgebra S o X is a collectio of subsets of X such that S is closed uder complemets ad coutable uios, that is if A S, the X \ A S, ad if (A ) is a sequece of sets i S, the A S, If the evelopig space X is clear, we use A to deote the complemet of A i X. It is easy to derive that a σalgebra is also closed uder the followig settheoretic operatios: Coutable itersectios we have A = A. Differeces we have A \ B = A B. Symmetric differeces we have A B =(A B) ( A B). Defiitio 5.2: Let (X, O) be a topological space. The collectio of Borel sets i X is the smallest σalgebra cotaiig the ope sets i O. Oe, of course, has to mae sure that this collectio actually exists. For this, ote that the itersectio of ay collectio of σalgebras is agai a σalgebra, so the Borel sets are just the itersectio of all σalgebras cotaiig O. (Note the the full power set of X is such a σalgebras, so we are ot taig a empty itersectio.) This defiitio of Borel sets is rather exteral. It does ot give us ay idea what Borel sets loo lie. Oe ca arrive at the family of Borel sets also through a costructio from withi. This reveals more structure ad gives rise to the Borel hierarchy. 5 1
2 The Borel hierarchy We will restrict ourselves from ow o to Polish spaces, to esure that every closed set is a coutable itersectio of ope sets (see exercises). To geerate the Borel sets, we start with the ope sets. By closig uder complemets, we obtai the closed sets. We also have to close uder coutable uios. The ope sets are already closed uder this operatio, but the closed sets are ot. Coutable uios of closed sets are classically ow as F σ sets. Their complemets, i.e. coutable itersectios of ope sets, are the G δ sets. We ca cotiue this way ad form the F σδ sets coutable itersectios of F σ sets the G δσ sets coutable uios of G δ sets ad so o. It is obvious that the σδotatio soo becomes rather impractical, ad hece we replace it by somethig much more coveiet, ad much more suggestive, as we will see later. Defiitio 5.3 (Borel sets of fiite order): Let X be a Polish space. We iductively defie the followig collectio of subsets of X. Σ 0 1 (X )={U: U X ope } Π 0 (X )={ A: A Σ0 (X )} = Σ0 (X ) Σ 0 +1 (X )={ A : A Π 0 (X )} Hece the ope sets are precisely the sets i Σ 0 1, the closed sets are the sets i Π 0 1, the F σ sets from the class Σ 0 2 etc. If it is clear what the uderlyig space X is, we drop the referece to it ad simply write Σ 0 ad Π0. Besides, we will say that a set A X is (or is ot) Σ 0 or Π0, respectively. Does the collectio of all Σ 0 ad Π0 exhaust the Borel sets of X? We will see that the aswer is o. We have to exted our iductive costructio ito the trasfiite ad cosider classes Σ 0, where ξ is a coutable ifiite ordial. ξ The Borel sets of fiite order We fix a Polish space X. We wat to establish the basic relatioships betwee the differet classes Σ 0 ad Π0 m for X. It is clear that Σ 0 1 Π0 1 ad Π0 1 Σ0 1. Furthermore, it follows from the defiitios that Π 0 Σ0 +1 ad Σ0 Π
3 Lemma 5.4: I a Polish metric space (X, d), every ope set is a F σ set. Proof. Let D = {x 1, x 2,...}X be a coutable dese subset, ad assume U X is ope. For ay >0, if δ<, the U δ (x) U (x) for ay x X. Let x i(1), x i(2),... ad 1, 2,... be such that U = U (x i() ). For each 1, let (δ () ) be such that δ() The U = <δ () +1 < < i, ad δ (). U () δ (x i() ). The set o the right had side is a coutable uio of closed sets. Corollary 5.5: Σ 0 1 Σ0 2 ad Π0 1 Π0 2. The secod statemet follows by passig to complemets: If F is closed, F = F = F = F, where the F are closed. There are also sets that ca be both Σ 0 2 ad Π0 2, but either Σ0 1 or Π0 1. For example, cosider the halfope iterval [0, 1). [0, 1)= [1, 1 1/]= ( 1/,1). Therefore, it maes sese to defie the hybrid classes 0 =Σ0 Π0. Usig iductio, we ca exted the iclusios i a straightforward way to higher. Theorem 5.6 (Wea Hierarchy Theorem): m Σ 0 1 Σ 0 2 Σ 0 3 Π 0 1 Π 0 2 Π
4 We also wat show that the iclusios are proper. For the first two levels, this ca be doe by explicit couterexamples. Ay coutable set is i Σ 0 2, sice a sigleto set is closed, ad a coutable set is a coutable uio of sigletos. However, there are coutable sets that are either ope or closed, e.g. {1/: 1}. The complemet is cosequetly a Π 0 2 set that is either ope or closed. Furthermore, the ratioals give a example of a Σ 0 2 set that is ot. This will be show later usig the cocept of Baire category. Π 0 2 It is much harder to fid specific examples for the higher levels, e.g. a Σ 0 5 set that is ot Σ 0 4. This separatio will be much facilitated by the itroductio of a logical/defiability framewor for the Borel sets. Therefore, we defer the proof for a while. Examples of Borel sets Cotiuity poits of fuctios Theorem 5.7 (Youg): Let f : X Y be a mappig betwee Polish spaces. The C f = {x : f is cotiuous at x} is a Π 0 2 (i.e. G δ) set. Proof. It is ot hard to see that f is cotiuous at a if ad oly if for ay >0, δ >0 x, y [x, y U δ (a) d( f (x), f ( y)) <]. ( ) Give >0, let C = {a: ( ) holds at a for }. We claim that C is ope. Suppose a C. Choose a suitable δ that witesses that a C. We show U δ (a) C. Let b U δ (a). Choose δ so that U δ (b) U δ (a). The x, y U δ (b) x, y U δ (a) d( f (x), f ( y)) <. Notice further that > implies C C. Hece we ca represet C f as a coutable itersectio of ope sets. C f = C 1/, 5 4
5 The fuctio f : give by 1 x irratioal, f (x)= 0 x = 0, 1/q x = p/q, p, q >0, p, q relatively prime is a fuctio that is cotiuous at every irratioal, discotiuous at every ratioal umber. As oted above, the ratioals are a Σ 0 2 set that is ot Π0 2. Hece there caot exist a fuctio g : that is discotiuous at exactly the irratioals. We fiish this lecture by showig that Youg s Theorem ca be reversed. Theorem 5.8: Give a Π 0 2 subset A of a perfect Polish space X, there exists a mappig f : X such that f is cotiuous at every poit i A, ad discotiuous at every other poit, i.e. C f = A. Proof. Fix a coutable dese subset D X. We first deal with the easier case that A is ope. Let 0 x A or x A D, f (x)= 1 otherwise. It is clear that f is cotiuous o A. Now assume x A. If x A, the there exists U (x) A. Ay U (x) U (x) cotais poits from both D ad D, so it is clear that f is ot cotiuous at x. Fially, let x A \ A. The f (x)=1, but poits of A are arbitrarily close, where f taes value 0. Now we exted this approach to geeral Π 0 2 sets. Suppose A = G, G ope. By replacig G with G = G 1 G, we ca assume that X = G 0 G 1 G 2 G 3... The idea is to defie f as above for each G ad the amalgamate the f is a suitable way. Assume for each, f : X is defied as above such that C f = G. Let (b ) be a sequece of positive real umbers such that for all, b > b, > 5 5
6 for example, b = 1/!. We ow form the series f (x)= b f (x). Sice f (x) 1, f (x) b <. Furthermore, ( f ) coverges uiformly to f, for f (x) f (x) b < b, > ad the last boud is idepedet of x ad coverges to 0. It follows by uiform covergece that if each f is cotiuous at x, f is cotiuous o x, too. Hece f is cotiuous o A. Now assume x A. The there exist such that x G \ G +1. Hece f 0 (x)= = f (x)=0. Agai, we distiguish two cases. First, assume x G +1. The there exists δ>0 such that U δ (x) G +1. This also implies U δ (x) G for ay + 1. Besides, sice G is ope, we ca chose δ sufficietly small so that U δ (x) G. For y D U δ (x) we have f ( y)=1 for all + 1, ad hece f ( y)= > b f ( y) > 0. O the other had, if y D U δ (x), the f ( y)=0 for all + 1, ad also f 0 ( y) = = f ( y) =0, sice y G, ad thus f ( y)=0. Hece there are poits arbitrarily close to x whose f values differ by a costat lower boud, which implies f is ot cotiuous i x. Fially, suppose x G +1. The f +1 (x) =1 ad hece f (x) b +1 > 0. O the other had, for ay y G +1, f ( y) >+1 b < b +1 = f (x). That is, there are poits arbitrarily close to x whose f value differs from f (x) by a costat lower boud. Hece f is discotiuous at x i this case, too. 5 6
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